$m(\text{Ca(OH)}_2) = \frac{0.03\, \text{г}}{74.1\, \text{г/моль}} = 0.000405\, \text{моль}$$
$m(\text{H}_2\text{O}) = \frac{25\, \text{см}^3}{1000\, \text{см}^3} = 0.025\, \text{л}$$
$M(\text{Ca(OH)}_2) = 40.08 + 2 \times 15.999 = 74.1\, \text{г/моль}$$
$n(\text{Ca(OH)}_2) = \frac{m(\text{Ca(OH)}_2)}{M(\text{Ca(OH)}_2)} = \frac{0.03}{74.1} = 0.000405\, \text{моль}$$
$n(\text{H}_2\text{O}) = \frac{m(\text{H}_2\text{O})}{M(\text{H}_2\text{O})} = \frac{0.025}{18.015} = 0.001387\, \text{моль}$$
$\text{мольные \, доли:} \, a(\text{Ca(OH)}_2) = \frac{n(\text{Ca(OH)}_2)}{n(\text{Ca(OH)}_2) + n(\text{H}_2\text{O})} = \frac{0.000405}{0.000405+0.001387} = \frac{0.000405}{0.001792} = 0.2261 \, \text{или} \, 22.61\%$$
Массовая доля гидроксида кальция в растворе равна 22,61%.
$
m(\text{Ca(OH)}_2) = \frac{0.03\, \text{г}}{74.1\, \text{г/моль}} = 0.000405\, \text{моль}
$$
$
m(\text{H}_2\text{O}) = \frac{25\, \text{см}^3}{1000\, \text{см}^3} = 0.025\, \text{л}
$$
$
M(\text{Ca(OH)}_2) = 40.08 + 2 \times 15.999 = 74.1\, \text{г/моль}
$$
$
n(\text{Ca(OH)}_2) = \frac{m(\text{Ca(OH)}_2)}{M(\text{Ca(OH)}_2)} = \frac{0.03}{74.1} = 0.000405\, \text{моль}
$$
$
n(\text{H}_2\text{O}) = \frac{m(\text{H}_2\text{O})}{M(\text{H}_2\text{O})} = \frac{0.025}{18.015} = 0.001387\, \text{моль}
$$
$
\text{мольные \, доли:} \, a(\text{Ca(OH)}_2) = \frac{n(\text{Ca(OH)}_2)}{n(\text{Ca(OH)}_2) + n(\text{H}_2\text{O})} = \frac{0.000405}{0.000405+0.001387} = \frac{0.000405}{0.001792} = 0.2261 \, \text{или} \, 22.61\%
$$
Массовая доля гидроксида кальция в растворе равна 22,61%.