A given sample of sea water is 2.925% (wt/wt). The molality of the solution is approximately:

5.2

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

0.52

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

0.94

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

1.02

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

Open in App

Solution

The correct option is B 0.52
$\frac{w t}{w t}$ percentage= $\frac{W e i g h t o f s o l u t e}{W e i g h t o f s o l u t i o n} \times 100$
$⟹ \frac{2.925}{100} = \frac{W e i g h t o f s o l u t e (W_{s})}{W_{s} + W e i g h t o f s o l v e n t}$
Considering one mole of water is taken, as it is sea water.
$⟹ 0.0295 = \frac{W_{s}}{W_{s} + 18}$
$⟹ (0.0295) (W_{s} + 18) = W_{s}$
$⟹ W_{s} (1 - 0.0295) = 18 (0.0295)$
$⟹ W_{s} (1 - 0.0295) = 0.531$
$⟹ W_{s} = 0.536$
Molality= $\frac{W_{s}}{G M W o f s o l u t e} \times \frac{1000}{W e i g h t o f s o l v e n t (g r a m s)} = \frac{0.536}{58.5} \times \frac{1000}{18} = \frac{536}{1053} = 0.52$

Suggest Corrections

Similar questions

15 % (w t . / v o l .)

H_{2} S O_{4}

1.1 g / c m^{3}

C_{2} H_{5} O H

- 10^{o} C

C_{2} H_{5} O H = 46, K_{f}

= 1.86 K k g m o l^{- 1}

26

d e n s i t y = 0.855

K_{f}

9

10

Join BYJU'S Learning Program