5 % of sulphur is present in insulin . the minimum molecular weight of insulin is
Answers
Answered by
0
mass by mass % of a solution=(mass of solute /mass of solution)×100
% of sulphur=5
5=32/x×100
x=3200/5
x=840
min. molecular mass of insulin is 870g
% of sulphur=5
5=32/x×100
x=3200/5
x=840
min. molecular mass of insulin is 870g
Similar questions