You visited us 2722 times! Enjoying our articles? Unlock Full Access!

Empirical & Molecular Formula

A certain non...

Question

A certain non-volatile electrolyte contains 40% carbon , 6.7% hydrogen and 53.3% oxygen . An aqueous solution containing 5% by mass of the solute boils at 100. $15^{\circ} C .$ Determine the molecular formula of the compound $(K_{b} = {0.5}^{\circ} C / m) :$

H C H O

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

C H_{3} O H

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

C_{2} H_{5} O H

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

C_{6} H_{12} O_{6}

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

Open in App

Solution

The correct option is D $C_{6} H_{12} O_{6}$
(d) Mole ration of $C : H : O$ is $1 : 2 : 1$ so empirical formula is $C H_{2} O$
$m = \frac{Δ T_{b}}{K_{b}} \Rightarrow \frac{0.15}{0.51} \Rightarrow 0.294$ ;
$0.294 = \frac{50}{M} \times \frac{1000}{950}; M ≃ 180$
$(C H_{2} O)_{n} = 180$ or $30 \times n = 180$ or $n = 6$ ;
$∴$ molecular formula is $C_{6} H_{12} O_{6}$ .

Suggest Corrections

Similar questions

101^{\circ} C .

(K_{f} = {1.86}^{\circ} C / m

K_{b} = {0.51}^{\circ} C / m)

288 g

C_{x} H_{2 x} O_{x}

90 g

{101.24}^{\circ} C

K_{b} (H_{2} O) = 0.512 K m o l^{- 1} k g, T_{b} (H_{2} O) = 100^{\circ} C

0.1 m

^{\circ} C

K_{b} = 0.52 k g m o l^{- 1} K

= 100^{\circ} C

40 %

6.7 %

53.3 %

60 g

Join BYJU'S Learning Program