Calculate the molality of a $1 L$ solution of 80% $H_{2} S O_{4}$ $(w / V)$ , given that the density of the solution is $1.80 {gmL}^{- 1}$ .

8.16 m

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

8.6 m

No worries! We‘ve got your back. Try BYJU‘S free classes today!

1.02 m

No worries! We‘ve got your back. Try BYJU‘S free classes today!

10.8 m

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A 8.16 m
Given:

$80 % o f H_{2} S O_{4} (w / v) m e a n s 80 g o f s o l u t e i n 100 m L o f s o l u t i o n$
weight of $H_{2} S O_{4}$ = 80 g
molecular weight of $H_{2} S O_{4}$ = 98 gmL
density of solution = $1.80 {gmL}^{- 1}$ .

$d e n s i t y o f s o l u t i o n = \frac{m a s s o f s o l u t i o n}{v o l u m e o f s o l u t i o n}$

$m a s s o f s o l u t i o n = 1.80 \times 1000 = 1800 g$
$w e i g h t o f H_{2} S O_{4} i n 1 L o f s o l u t i o n = \frac{80}{100} \times 1000 = 800 g$

$w e i g h t o f s o l v e n t i n s o l u t i o n = 1800 - 800 g = 1000 g = 1 k g$

$m o l a l i t y o f s o l u t i o n = \frac{m a s s o f s o l u t e}{m o l e c u l a r m a s s o f s o l u t e \times m a s s o f s o l v e n t (k g)}$

$= \frac{800 g}{98 {gmL}^{- 1} \times 1 k g} = 8.16 m$ .

Suggest Corrections

Similar questions

1 L

80 % H_{2} S O_{4} (w / v)

1.80 g m l^{- 1}

1 L

H_{2} S O_{4}

(w / V)

1.80 {gmL}^{- 1}

1 L

93 % H_{2} S O_{4}

1.84 g m L^{- 1}

Join BYJU'S Learning Program