bài 4:
a,500ml=0.5l
$C_{M}$= $\frac{n}{V}$
hay 2=$\frac{n}{0.5}$
=>n=2*0.5=1(mol)
=>mKNO3=1*(39+14+16*3)=101(g)
b,$C_{M}$= $\frac{n}{V}$
hay 0.3=$\frac{n}{2}$
=>n=0.3*2=0.6(mol)
=>mNa2SO4=0.6*(23*2+32+16*4)=85.2(g)
bài 5
ta có 36(g)muối hòa tan được trong 100(g) H2O=>dd bão hòa
=>mdd=36+100=136(g)
=>C%=$\frac{36}{136}$*100=26.4%