Hóa 8

C

chonhoi110

Gọi $m_{CuSO_4. 5H_2O }=a$ (g)

$\Longrightarrow n _{CuSO_4}=n_{CuSO_4. 5H_2O }=\dfrac{a}{250}$ (mol)

$\Longrightarrow m_{CuSO_4}=\dfrac{a}{250}.160=0,64a$ (g)

$m_{CuSO_4-trong-dd-tạo-ra}=400.2\%+0,64a=8+0,64a$

$m_{dd-tạo-ra}=400+a$

$C\%_{ddCuSO_4}=\dfrac{160.1}{1,1.10}\%=\dfrac{160}{11}\%$

$\Longrightarrow \dfrac{8+0,64a}{400+a}=\dfrac{160}{11}\%$

$\Longrightarrow a=101,47$ (g)
 
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