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Molarity

If 100 ml of ...

Question

If 100 ml of $1 M H_{2} S O_{4}$ solution is mixed with 100 ml of $9.8 %$ (w/w) $H_{2} S O_{4}$ solution ( $d = 1 g / m l$ ) then:

Molarity of resulting solution is 1

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Volume of solution becomes 200 ml

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Mass of H_{2} S O_{4} in the solution is 98 g

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Mass of H_{2} S O_{4} in the solution is 19.6 g

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Solution

The correct option is D $Mass of H_{2} S O_{4} in the solution is 19.6 g$
$w = 98 g$
$M = 98 g$
$d = 1 g / m l$
we know the formula of Molarity is
$M = \frac{n u m b e r o f m o l e s o f s o l u t e}{V o l u m e o f s o l u t i o n (i n l i t e r)}$
and number of mole is given as
$n = \frac{m a s s o f s o l u t e}{m o l a r m a s s o f s o l u t e}$
(A) $M_{1} = 1 M V_{1} = \frac{100}{1000} = 0.1 L$
$9.8 g$ of $H_{2} S O_{4}$ present in 100 g of solution.
Since, $d = 1 g m l^{- 1} ∴ V_{2} = 100 m l = 0.1 L$
Sinec, $9.8 g$ is in $100 m l$ of solution, So, $98 g$ will be present in $1000 m l$ or $1 L$ of solution.
Molarity $M_{2} = \frac{Moles of solute}{Volume} M_{2} = \frac{98}{98 \times 1} = 1 M$

$M_{3} = \frac{M_{1} V_{1} + M_{2} V_{2}}{V_{1} + V_{2}} M_{3} = \frac{(1 \times 0.1) + (1 \times 0.1)}{0.2} = 1 M$

(B) Volume of resulting solution $= 100 + 100 = 200$ ml
(D) Using molarity formula,
$Mass of H_{2} S O_{4} = \frac{200 \times 1}{1000} \times 98 = 19.6$ g.

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