$m_{H_2SO_4}=14,7.10\%=1,47g$
Sau khi thêm $CuSO_4.xH_2O$, $C\%_{H_2SO_4}=10-2,65=7,35\%$
$\to m_X=1,47:7,35\%=20g$
$\to a=20-14,7=5,3g$
Nếu cho $20g$ $X$ tác dụng với $Ba(OH)_2$ dư, $m_{\downarrow}=10,5122g$
Đặt $n_{CuSO_4.xH_2O}=b(mol)$
$n_{H_2SO_4}=\dfrac{1,47}{98}=0,015(mol)$
$Ba(OH)_2+H_2SO_4\to BaSO_4+2H_2O$
$CuSO_4+Ba(OH)_2\to BaSO_4+Cu(OH)_2$
$n_{BaSO_4}=0,015+b(mol)$
$n_{Cu(OH)_2}=b(mol)$
$\to 98b+233(0,015+b)=10,5122$
$\to b=0,0212(mol)$
$M_{CuSO_4.xH_2O}=\dfrac{5,3}{0,0212}=250=160+18x$
$\to x=5$
Vậy CTPT tinh thể muối ngậm nước là $CuSO_4.5H_2O$