The molal elevation constant for water is 0.52 . What will be the boiling point of 2 molar sucrose solution at 1atm pressure? (Assume B.P. of pure water as 100oC )

Question

The molal elevation constant for water is 0.52 . What will be the boiling point of 2 molar sucrose solution at 1atm pressure? (Assume B.P. of pure water as 100oC ) (A)  101.04oC  (B)  100.26oC  (C)  100.52oC  (D)  99.74oC

Tardigrade Admin · Accepted Answer

Δ Tb=kb × molality for dilute solution molarity = molarity = 2 (given) and kb = 0.52(given) ∴ Δ Tb = 0.52 × 2 = 1.04° C Now, Δ Tb = B. P. of solution - B.P. pf H2O = 1.04 + 100 = 101.04 ° C

Q. The molal elevation constant for water is $0.52$ . What will be the boiling point of $2$ molar sucrose solution at $1 a t m$ pressure? (Assume B.P. of pure water as $100^{o} C$ )

$101.04^{o} C$

$100.26^{o} C$

$100.52^{o} C$

$99.74^{o} C$

$Δ T_{b} = k_{b} \times$ molality
for dilute solution
molarity = molarity $= 2$ (given)
and $k_{b} = 0.52$ (given)
$∴ Δ T_{b} = 0.52 \times 2 = 1.0 4^{\circ} C$
Now,
$Δ T_{b} =$ B. P. of solution - B.P. pf $H_{2} O$
$= 1.04 + 100 = 101.0 4^{\circ} C$

Q. The molal elevation constant for water is 0.52 . What will be the boiling point of 2 molar sucrose solution at 1atm pressure? (Assume B.P. of pure water as 100oC )

101.04oC

100.26oC

100.52oC

99.74oC