Question

在20℃时,把30克硝酸钠溶解在60克水里,所得溶液为不饱和溶液.为了使它成为饱和溶液,可以用下面的方法（20℃使硝酸钠

Answer 1

NaNO3 H2O

88 100

30 x

y 60

1.蒸发溶剂法.88:100 = 30 :x

x = 34.1 g

需蒸发掉水:60 - 34.1 = 25.9 g

2.增加溶质.88:100 = y :60

y = 52.8 g

增加溶质:52.8 - 30 = 17.2 g

NaNO3 NaNO3 + H2O (20℃)

88 88 + 100 = 188

x 263.2

88:188 = x :263.2

x = 123.2 g

H2O:263.2 - 123.2 = 140 g

NaNO3 H2O (60℃)

110 100

123.2 a

b 140

110 :100 = 123.2 :a

a = 112 g

蒸发水:140 - 112 = 28 g

110 :100 = b :140

b = 154 g

增加溶质:154 - 123.2 = 30.8 g