Sau khi trộn, $m_{dd}=150+45=195g$
$\to m_{K_2CO_3\text{sau khi trộn}}=195.15\%=29,25g$
$m_{K_2CO_3\text{dd $10\%$}}=150.10\%=15g$
$\to n_{K_2CO_3\text{tinh thể}}=\dfrac{29,25-15}{138}=0,10326(mol)$
$\to n_{H_2O\text{tinh thể}}=\dfrac{45-0,10326.138}{18}=1,70834(mol)$
$\to x=\dfrac{1,70834}{0,10326}=16,5$