Question

将11.5g钠投入到89g水中（足量）,充分反应后所得溶液密度为1.25g/m3,则计算：（1）所得溶液中的质量分数是多

Answer 1

(1) n(Na）=n(NaOH)=11.5g/23g/mol=0.5mol

n(H2)=1/2 * 0.5mol =0.25mol

m(H2)=0.25mol *2g/mol =0.5g

m(NaOH)=0.5*40=20g

m(溶液）=11.5+89-0.5=100g

w(NaOH)%=20g/100g *100%=20%

(2)V(溶液）=100g / 1.25g/m3=80mL=0.08L

c(NaOH)=0.5mol / 0.08L=6.22mol/L