将40%的硫酸溶液50g，倒入一种未知溶质的质量分数的硫酸溶液20g中，混合后加水70g稀释．取出稀释后的溶液10g，恰好跟16%的氢氧化钠溶液10g完全中和．则未知溶质质量分数的硫酸溶液的溶质

Question

将40%的硫酸溶液50g，倒入一种未知溶质的质量分数的硫酸溶液20g中，混合后加水70g稀释．取出稀释后的溶液10g，恰好跟16%的氢氧化钠溶液10g完全中和．则未知溶质质量分数的硫酸溶液的溶质质量分数是______．

Answer 1

设与氢氧化钠反应的硫酸的质量为x
2NaOH+H₂2SO₄4═Na₂2SO₄4+2H₂2O
80          98
10g×16%    x
(80)/(10g×16%)=(98)/(x)
         x=1.96g
混合后溶液的溶质质量分数为：(1.96g)/(10g)×100%=19.6%
设未知溶质质量分数的硫酸溶液的溶质质量分数是y
(50g×40%+20g×y)/(50g+20g+70g)×100%=19.6%
                      y=37.2%
答：未知溶质质量分数的硫酸溶液的溶质质量分数是37.2%． (80)/(10g×16%)80808010g×16%10g×16%10g×16%=(98)/(x)
         x=1.96g
混合后溶液的溶质质量分数为：(1.96g)/(10g)×100%=19.6%
设未知溶质质量分数的硫酸溶液的溶质质量分数是y
(50g×40%+20g×y)/(50g+20g+70g)×100%=19.6%
                      y=37.2%
答：未知溶质质量分数的硫酸溶液的溶质质量分数是37.2%． (98)/(x)989898xxx
         x=1.96g
混合后溶液的溶质质量分数为：(1.96g)/(10g)×100%=19.6%
设未知溶质质量分数的硫酸溶液的溶质质量分数是y
(50g×40%+20g×y)/(50g+20g+70g)×100%=19.6%
                      y=37.2%
答：未知溶质质量分数的硫酸溶液的溶质质量分数是37.2%． (1.96g)/(10g)1.96g1.96g1.96g10g10g10g×100%=19.6%
设未知溶质质量分数的硫酸溶液的溶质质量分数是y
(50g×40%+20g×y)/(50g+20g+70g)×100%=19.6%
                      y=37.2%
答：未知溶质质量分数的硫酸溶液的溶质质量分数是37.2%． (50g×40%+20g×y)/(50g+20g+70g)50g×40%+20g×y50g×40%+20g×y50g×40%+20g×y50g+20g+70g50g+20g+70g50g+20g+70g×100%=19.6%
                      y=37.2%
答：未知溶质质量分数的硫酸溶液的溶质质量分数是37.2%．