Question

On heating $5\times {10}^{-3}$ equivalent of CaCl${}_2$.xH${}_2$O, 0.18 gm of water is obtained. What is the value of x?

Answer 1

No. of equivalents $=5\times {10}^{-3}.$

Equivalent weight of $CaC{l}_2.x{H}_{2}O=55.5+9x$

Given weight $=0.18gm$

No. of gram equivalen $=\frac{\text{given weight}}{\text{equivalent weight of}CaC{l}_2.x{H}_{2}O}$

$5\times 10-3=\frac{0.13}{55.5+9x}9x=55.5-36x=\frac{19.5}{9}\approx 2$