How many moles of sucrose should be dissolved it 500 g of water so as to get a solution which has 3 difference of $104 ° C$ between boiling point and freezing point? $(K_{f} = 1.86 K kg {mol}^{- 1}, K_{b} = 0.52 K kg m o l^{- 1})$

see full answer

High-Paying Jobs That Even AI Can’t Replace — Through JEE/NEET

🎯 Hear from the experts why preparing for JEE/NEET today sets you up for future-proof, high-income careers tomorrow.

An Intiative by Sri Chaitanya

1.68

8.40

1.492

0.840

answer is D.

(Unlock A.I Detailed Solution for FREE)

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

We know that the boiling point and freezing point of water are $100 ° C$ and $0 ° C$ respectively, So

$104^{0} C = {ΔT}_{b} - {ΔT}_{f} + 100^{\circ} C$

$\begin{array}{l} (Δ T_{b} - Δ T_{f}) = 4 \\ (1.86 + 0.52) \times \frac{n}{500} \times 1000 = 4 \Rightarrow n = 0.84 0 \end{array}$

$=$ 0.840 moles of sucrose.

Watch 3-min video & get full concept clarity

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test