SATHEE: Some Basic Concepts Of Chemistry Question 70

Some Basic Concepts Of Chemistry Question 70

Question: Specific volume of cylindrical virus particle is $6.02 \times 10^{- 2} c c / g .$ whose radius and length are 7 $\overset{o}{A},$ & 10 $\overset{o}{A},$ respectively.

If $N_{A} = 6.02 \times 10^{23} m o l^{- 1}$ find molecular weight of virus

Options:

A) $3.08 \times 10^{3} k g / m o l$

B) $3.08 \times 10^{4} k g / m o l$

C) $1.54 \times 10^{4} k g / m o l$

D) $15.4 k g / m o l$

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Answer:

Correct Answer: D

Solution:

[d] Specific volume (volume of 1 g) of cylindrical virus particle $= 6.02 \times 10^{- 2} c c / g$ Radius of virus $= 7 \overset{o}{A}, = 7 \times 10^{- 8} c m$ Length of virus $= 10 \times 10^{- 8} c m$ Volume of virus = $π r^{2} I = \frac{22}{7} \times {(7 \times 10^{- 8})}^{2} \times 10 \times {10^{-}}^{8}$ $= 154 \times 10^{- 23} c c$

$W t . o f o n e v i r u s p a r t i c l e = \frac{v o l u m e}{s p e c i f i c v o l u m e}$

$∴$ Mol. wt. of virus = Wt. of $N_{A}$ particle $= \frac{154 \times 10^{- 23}}{6.02 \times 10^{- 2}} \times 6.02 \times 10^{23} = 15400 g / m o l$

$= 15.4 k g / m o l$

Some Basic Concepts Of Chemistry Question 70

Question: Specific volume of cylindrical virus particle is $6.02 \times 10^{- 2} c c / g .$ whose radius and length are 7 $\overset{o}{A},$ & 10 $\overset{o}{A},$ respectively.

Options:

Answer:

Solution:

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Some Basic Concepts Of Chemistry Question 70

Question: Specific volume of cylindrical virus particle is 6.02×10−2cc/g. whose radius and length are 7 Ao, & 10 Ao, respectively.

Options:

Answer:

Solution:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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Question: Specific volume of cylindrical virus particle is $6.02 \times 10^{- 2} c c / g .$ whose radius and length are 7 $\overset{o}{A},$ & 10 $\overset{o}{A},$ respectively.