SATHEE: Some Basic Concepts of Chemistry

Some Basic Concepts of Chemistry - Result Question 42

####42. What is the mass of precipitate formed when 50 $m L$ of $16.9$ solution of $A g N O_{3}$ is mixed with 50 $m L$ of $5.8$ solution? [2015 RS] $(A g = 107.8, N = 14, O = 16, N a = 23, C l = 35.5)$

(a) $28 g$

(b) $3.5 g$

(d) $14 g$

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Solution:

$\begin{aligned} (\frac{16.9}{100} \times 50) = 8.45 g o f^{2 g N O_{3}} \\ n_{mole} = \frac{8.45 g}{(107.8 + 14 + 16 \times 3) g / m o l} \\ = (\frac{8.45 g}{169.8 g / m o l}) = 0.0497 m o l e s \end{aligned}$

$50 m L$ of $5.8$ solution of $N a C l$ contain

$N a C l = (\frac{5.8}{100} \times 50) = 2.9 g$ $n_{N a C l} = \frac{2.9 g}{(23 + 35.5) g / m o l} = 0.0495 m o l e s$

$A g N O_{3} + N a C l \to A g C l ↓ + N a^{+} + N O_{3}^{-}$

$1 mole 1 mole 1 mole$

$∴ 0.049 mole 0.049 mole 0.049 mole of A g C l$

$n = \frac{w}{M} \to w = (n_{AgCl}) \times$ Molecular Mass

$= (0.049) \times (107.8 + 35.5) = 7.02 g$

Some Basic Concepts of Chemistry - Result Question 42

Solution:

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Some Basic Concepts of Chemistry - Result Question 42

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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