<h3 id="successchemistry-solution-1success"><Success>Chemistry solution-1</Success></h3> <div style='float: left; width:50%;font-size:30px; text-align:left'> </div> <div style='float: right; width:0%;'> <p><img alt="image" src="https://temp-public-img-folder.s3.ap-south-1.amazonaws.com/sathee.prutor.images/subject-images/iitpal/image/chemistry-class-12-unit-02-chapter-01-solutions-l-1-4-1.jpg"></p>

<h3 id="successhomogenous-solutionsuccess"><Success>Homogenous solution</Success></h3> <div style='float: left; width:100%;font-size:30px; text-align:left'> <p>Large quantity - solvent</p> <p>Other components - solutes</p> <p>Composite-</p> </div> <div style='float: right; width:0%;'> <p><img alt="image" src="https://temp-public-img-folder.s3.ap-south-1.amazonaws.com/sathee.prutor.images/subject-images/iitpal/image/chemistry-class-12-unit-02-chapter-01-solutions-l-1-4-2.jpg"></p>

<h3 id="successmass-percentagesuccess"><Success>Mass percentage</Success></h3> <div style='float: left; width:100%;font-size:30px; text-align:left'> <p><strong>Mass percentage</strong> $=\frac{\text{ mass of the component in the solution}}{\text {Total mass of the solution}} \times 100$</p> <ul> <li>10% glucose in water <ul> <li>10 g of glucose + 90 g of water = 100g 10% aquous glucose solution</li> </ul> </li> </ul> </div> <div style='float: right; width:1%;'> <p><img alt="image" src="https://temp-public-img-folder.s3.ap-south-1.amazonaws.com/sathee.prutor.images/subject-images/iitpal/image/chemistry-class-12-unit-02-chapter-01-solutions-l-1-4-3.jpg"></p> </div>

<h3 id="successvolume-percentagesuccess"><Success>Volume percentage</Success></h3> <div style='float: left; width:100%;font-size:30px; text-align:left'> <p><strong>Volume percentage</strong> = $ \frac {\text {Volume of the component in the solution}}{\text {Total volume of the solution}} \times 100$</p> <ul> <li>10% ethanol solution in water <ul> <li>10ml of ethanol + enough water = 100ml of solution</li> </ul> </li> </ul> </div> <div style='float: right; width:0%;'> <p><img alt="image" src="https://temp-public-img-folder.s3.ap-south-1.amazonaws.com/sathee.prutor.images/subject-images/iitpal/image/chemistry-class-12-unit-02-chapter-01-solutions-l-1-4-4.jpg"></p> </div>

<h3 id="successmass-by-volumesuccess"><Success>Mass by volume</Success></h3> <div style='float: left; width:100%;font-size:30px; text-align:left'> <p><strong>Mass by volume</strong> $=\frac{\text{ mass of the component in the solution}}{\text {Total volume of the solution}} \times 100$</p> <ul> <li>10% ethanol solution in water <ul> <li>10ml of ethanol + enough water = 100ml of solution</li> </ul> </li> </ul> </div> <div style='float: right; width:0%;'> <p><img alt="image" src="https://temp-public-img-folder.s3.ap-south-1.amazonaws.com/sathee.prutor.images/subject-images/iitpal/image/chemistry-class-12-unit-02-chapter-01-solutions-l-1-4-5.jpg"></p> </div>

<h3 id="successppmpart-per-millionsuccess"><Success>PPM(part per million)</Success></h3> <div style='float: left; width:50%;font-size:30px; text-align:left'> <p><strong>PPM(part per million)</strong></p> <p>= $\frac {\text {parts of the component}}{\text {Total parts}} \times 10^6$</p> <p>$ 6 \times 10^{-3}g$ at $O_2$ in 1L water (1030g)</p> <p>$\frac {6 \times 10^{-3}}{1030g} \times 10^6 = 5.8 PPM$</p> <p>= 5.8g of $O_2 10^6g$ water</p> <p>= 5.8$\mu$ g of $O_2$ 1g water</p> </div> <div style='float: right; width:50%;'> <p><img alt="image" src="https://temp-public-img-folder.s3.ap-south-1.amazonaws.com/sathee.prutor.images/subject-images/iitpal/image/chemistry-class-12-unit-02-chapter-01-solutions-l-1-4-6.jpg"></p>

<div style='float: left; width:50%;font-size:30px; text-align:left'> <p>mole fraction</p> <p>$ X_A=\frac{n_A}{n_A+n_B} \quad X_B = \frac{n_B}{n_A+n_B}$</p> <p>$ X_A+X_B=1$</p> <p>$ X_l=\frac{n_l}{\sum_j n_j}$</p> <p>$ \sum_l X_l=12$</p> </div> <div style='float: right; width:50%;'> <p><img alt="image" src="https://temp-public-img-folder.s3.ap-south-1.amazonaws.com/sathee.prutor.images/subject-images/iitpal/image/chemistry-class-12-unit-02-chapter-01-solutions-l-1-4-7.jpg"> <img alt="image" src="https://temp-public-img-folder.s3.ap-south-1.amazonaws.com/sathee.prutor.images/subject-images/iitpal/image/chemistry-class-12-unit-02-chapter-01-solutions-l-1-4-7.a.jpg"></p>

<div style='float: left; width:50%;font-size:30px; text-align:left'> <p>molarity = $\frac {\text {moles of solute}}{\text {volume of solution (litre)}}$</p> <p>0.25m NaCl solution in water</p> <p>0.25 moles of NaCl in 1L solution</p> <p>0.25 moles</p> <p>molality = $\frac {\text {moles of solute}}{\text {weight of solvent (kg)}}$</p> <p>1m KBr solution in water</p> <p>1mole of KBr in 1kg of water</p> </div> <div style='float: right; width:50%;'> <p><img alt="image" src="https://temp-public-img-folder.s3.ap-south-1.amazonaws.com/sathee.prutor.images/subject-images/iitpal/image/chemistry-class-12-unit-02-chapter-01-solutions-l-1-4-8.jpg"></p>

<div style='float: left; width:50%;font-size:30px; text-align:left'> <p>mass parcentage = $\frac {\text{mass of the component}}{\text{Total mass}} \times 100$</p> <p>$=\frac{22}{122+22} \times 100=15.3$</p> <p>$100-153=84.7 %$</p> <p>$\frac{122}{144} \times 100$</p> <p>$C_6 H_6(72+6=78) \quad =\frac {22g}{78} = n_1 = 0.28$</p> <p>$CCl_4(12+35.5 \times 4 = 154) \quad W_2= \frac {122g}{154} = n_2$</p> <p>$\begin{aligned} & X_1=\frac{n_1}{n_1+n_2}=\frac{0.28}{0.28+0.79}= 0.26 \\ & X_2=1-X_1=74 \ & \end{aligned}$</p> </div> <div style='float: right; width:50%;'> <p><img alt="image" src="https://temp-public-img-folder.s3.ap-south-1.amazonaws.com/sathee.prutor.images/subject-images/iitpal/image/chemistry-class-12-unit-02-chapter-01-solutions-l-1-4-9.jpg"></p>

<div style='float: left; width:50%;font-size:30px; text-align:left'> <p>molarity = $\frac{\text{moles of solute}}{\text{volume of solution (L)}}$</p> <p>a. $\frac {30}{290.93}g \quad Co(NO_3) . 6H_2O \quad \frac{0.103}{4.3} \quad 0.024M$</p> <p>$V = 4.3L$</p> <p>$n_2 = 0.103$</p> </div> <div style='float: right; width:50%;'> <p><img alt="image" src="https://temp-public-img-folder.s3.ap-south-1.amazonaws.com/sathee.prutor.images/subject-images/iitpal/image/chemistry-class-12-unit-02-chapter-01-solutions-l-1-4-10.jpg"></p>

<div style='float: left; width:50%;font-size:30px; text-align:left'> <p>molarity = $ \frac {\text{moles of solute}}{\text{volume of solution(Litre)}} \quad =\frac {0.5 \times 30}{1000 \times \frac {500}{1000}}$</p> <p>a. $30ml 0.5 M H_2SO_4 \rightarrow 500ml$</p> <p>$0.5 = \frac {\text{moles of solute}}{30/1000}$</p> <p>moles of solute = $ \frac {0.5 \times 30}{1000}$</p> </div> <div style='float: right; width:50%;'> <p><img alt="image" src="https://temp-public-img-folder.s3.ap-south-1.amazonaws.com/sathee.prutor.images/subject-images/iitpal/image/chemistry-class-12-unit-02-chapter-01-solutions-l-1-4-11.jpg"></p>

<div style='float: left; width:50%;font-size:30px; text-align:left'> <p>molality = $\frac {\text{moles of solute}}{\text {Wt of solvent}}$</p> <p>0.25 =</p> <p>2.5kg of solution</p> <p>0.25m</p> <p>$Urea(NH_2CO NH_2) = 16 + 12 + 16 + 16 = 60g$</p> <p>moles of solute = $\frac {4 \times 1000}{60}$</p> <p>W $\quad$ Weight of solute (kg)</p> <p>S - W Solvent</p> <p>molality = $\frac {\text{moles of solute}}{\text {Wt of solvent}}$</p> <p>$0.25=\frac{w \times 1000}{60(2.5-\omega)}$</p> <p>$\frac{W}{25-W}=\frac{0.25 \times 6 \phi}{100 \phi}=1.5 \times 10^{-2}=\frac{.015}{1}$</p> <p>$\frac{w}{25-w+w}=\frac{0.015}{1+0.015} \Rightarrow h=\frac{0.015}{1015} \times 2.5$</p> <p>$=0.037 \mathrm{Kg}$</p> <p>$=37g$</p> </div> <div style='float: right; width:50%;'> <p><img alt="image" src="https://temp-public-img-folder.s3.ap-south-1.amazonaws.com/sathee.prutor.images/subject-images/iitpal/image/chemistry-class-12-unit-02-chapter-01-solutions-l-1-4-12.jpg"> <img alt="image" src="https://temp-public-img-folder.s3.ap-south-1.amazonaws.com/sathee.prutor.images/subject-images/iitpal/image/chemistry-class-12-unit-02-chapter-01-solutions-l-1-4-12.a.jpg"> <img alt="image" src="https://temp-public-img-folder.s3.ap-south-1.amazonaws.com/sathee.prutor.images/subject-images/iitpal/image/chemistry-class-12-unit-02-chapter-01-solutions-l-1-4-12.b.jpg"></p>

<div style='float: left; width:50%;font-size:30px; text-align:left'> <p>a. molality</p> <p>b. molarity</p> <p>c. mole fraction</p> <p>20% mass/mass $\quad$ f = 1.202g/ml</p> <p>20g $\quad$ K I(39.1 126q) = 1660g/mol</p> <p>80g $\quad$ solvent water</p> <p>$^nKI = \frac{20}{166}$ = 0.12moles</p> <p>molality = $\frac {\text{moles of solvent}}{\text{water of solvent}}$</p> <p>$= \frac {0.12}{80} \times 1000 = \frac {120}{80} = 1.5m$</p> <p>molarity = $\frac {\text{moles of solute}}{\text{volume of solution}}$</p> <p>= $\frac {0.12}{83.19} \times 1000 = 1.44$</p> <p>V = $\frac {100g}{1.202g/ml} = 83.19ml$</p> <p>$^XKI = \frac {4^nKI}{^nKI + ^nH_2O} = \frac {0.12}{0.12+ \frac {80}{18}} = 0.026$</p> </div> <div style='float: right; width:50%;'> <p><img alt="image" src="https://temp-public-img-folder.s3.ap-south-1.amazonaws.com/sathee.prutor.images/subject-images/iitpal/image/chemistry-class-12-unit-02-chapter-01-solutions-l-1-4-13.jpg"> <img alt="image" src="https://temp-public-img-folder.s3.ap-south-1.amazonaws.com/sathee.prutor.images/subject-images/iitpal/image/chemistry-class-12-unit-02-chapter-01-solutions-l-1-4-13.a.jpg"></p>

<div style='float: left; width:50%;font-size:30px; text-align:left'> <p>$300\mathrm{g} \text { of } 25 %$ $+400 \mathrm{g}\text { of } 40 %$</p> <p>$25 \times 300 / 100=75 \mathrm{g}$</p> <p>$400 \times 40 / 100=160 \mathrm{g}$</p> <p>Mass percentage $=\frac{\text { mass of the component }}{\text { Total mass }}$</p> <p>$=\frac{235}{700} \times 100=33.6 %$</p> </div> <div style='float: right; width:50%;'> <p><img alt="image" src="https://temp-public-img-folder.s3.ap-south-1.amazonaws.com/sathee.prutor.images/subject-images/iitpal/image/chemistry-class-12-unit-02-chapter-01-solutions-l-1-4-14.jpg"></p>

<div style='float: left; width:50%;font-size:30px; text-align:left'> <ol> <li>molality = $ \frac {\text {moles of solute}}{\text{Wt of solvent}}$</li> </ol> <p>= 3.59/.2 = 17.95m</p> <p>3.95mls = $ \frac {222.6g}{62g/mol}$ ethylene glycol</p> <p>200g of water</p> <p>$\rho = 1.072g/ml$</p> <p>V = $\frac {422.6g}{1.072g/ml}$</p> <p>= 394.2ml</p> <p>molarity =$\frac {\text{moles of solute}}{\text {volume of solution}}$</p> <p>= $\frac{3.59}{3942}$</p> <p>= 9.11 M</p> </div> <div style='float: right; width:50%;'> <p><img alt="image" src="https://temp-public-img-folder.s3.ap-south-1.amazonaws.com/sathee.prutor.images/subject-images/iitpal/image/chemistry-class-12-unit-02-chapter-01-solutions-l-1-4-15.jpg"> <img alt="image" src="https://temp-public-img-folder.s3.ap-south-1.amazonaws.com/sathee.prutor.images/subject-images/iitpal/image/chemistry-class-12-unit-02-chapter-01-solutions-l-1-4-15.a.jpg"></p>

<h3 id="successthank-yousuccess"><Success>Thank you</Success></h3> <div style='float: left; width:50%;font-size:30px; text-align:left'> </div> <div style='float: right; width:50%;'> <!---->