Question

Element M is a shiny and highly reactive metal and element X is a highly reactive non metal. They react to form a compound MX, a colourless brittle white solid. MX conducts electricity in aqueous solution and in molten state. When chlorine gas is bubled through an aqueous solution, containing MX, a reddish-brown liquid appears and ${Cl}^{-}$ ions are formed.
$2MX+{Cl}_2⟶2MCl+X_2$
$X_2$ is decolourised by unsaturated hydrocarbon and aq solution of MX containing 1.19 g MX in 100g $H_{2}O$ freezes at ${-3.72}^{0}C$. Molar depression constant of $H_{2}O$ is ${1.86}^0$ ${mol}^{-1}$kg. M is potassium.
if true enter 1, else enter 0.

Answer 1

$X_2$ is decolourised by unsaturated hydrocarbon
thus $X_2$ is ${Br}_2$
and MX is MBr.
Depression in freezing point
$\left(\Delta T_f\right)=0-\left({-3.72}^o\right)={3.72}^o$
$\Delta T_f=\frac{1000K_f{w}_1\left(i\right)}{m_1{w}_1}$
$m_1$ (molar mass of $MBr$)=?
$K_f$ (molar depression constant)=${1.86}^o{mol}^{-1}kg$
$w_1$ (mass of MX)=11.9g
$w_2$ (mass of $H_{2}O$)=100g
i(van't Hoff factor)=$(1+x)=1+1=2$
$x$ = degree of ionisation = 1
(assuming MX is conducting and 100% ionised.)
$\therefore m_1=\frac{1000\times 1.86\times 11.9\times 2}{3.72\times 100}$
=119g ${mol}^{-1}$
$\therefore Thus,M+Br=119g{mol}^{-1}$
$\therefore M+80=119$
M=39
Thus, M is potassium.