Equimolal aqueous solutions of NaCl and BaCl2 are prepared. If the freezing point of NaCl is -2°C , the freezing point of BaCl2 solution is expected to be

Question

Equimolal aqueous solutions of NaCl and BaCl2 are prepared. If the freezing point of NaCl is -2°C , the freezing point of BaCl2 solution is expected to be (A)  -2°C  (B)  -3°C  (C)  -1.5°C  (D)  -1.66°C

Tardigrade Admin · Accepted Answer

i=N a C l=2, i for B a C l2=3 Hence, (Δ Tf(N a C l)/Δ Tf(B a C l2))=(2/3) or Tf(B a C l2)=(3/2) × 2=3° .(Δ Tf N a C l)=2) So that Tf for Ba B a C l2=-3° C

Q. Equimolal aqueous solutions of $N a Cl$ and $B a C l_{2}$ are prepared. If the freezing point of $N a Cl$ is $- 2^{\circ} C$ , the freezing point of $B a C l_{2}$ solution is expected to be

$- 2^{\circ} C$

$- 3^{\circ} C$

$- 1. 5^{\circ} C$

$- 1.6 6^{\circ} C$

$i = N a Cl = 2$ ,
$i$ for $B a C l_{2} = 3$
Hence, $\frac{Δ T _{f} ( N a Cl )}{Δ T _{f} ( B a C l _{2} )} = \frac{2}{3}$
or $T_{f} (B a C l_{2}) = \frac{3}{2} \times 2 = 3^{\circ}$
$(Δ T_{f} N a Cl) = 2)$
So that $T_{f}$ for $B a$
$B a C l_{2} = - 3^{\circ} C$

Q. Equimolal aqueous solutions of NaCl and BaCl2​ are prepared. If the freezing point of NaCl is −2∘C , the freezing point of BaCl2​ solution is expected to be

−2∘C

−3∘C

−1.5∘C

−1.66∘C