b) Từ phần a $=> n_{HCl}=1,8 (mol)=>m_{HCl}=1,8.(1+35,5)=65,7(g)$$=>C $ %$=\frac{m_{HCl}}{m_{dd HCl}}.100$%$=36$(%)$<=>m_{dd HCl}=182,5(g)$Mà : $m=D.V=>V_{dd HCl}=182,5/1,18\approx 154,66(ml)$
b) Từ phần a $=> n_{HCl}=1,8 (mol)=>m_{HCl}=1,8.(1+35,5)=65,7(g)$$=>C $ %$=\frac{m_{HCl}}{m_{dd HCl}}.100$%$=36$%$<=>m_{dd HCl}=182,5(g)$Mà : $m=D.V=>V_{dd HCl}=182,5/1,18\approx 154,66(ml)$
b) Từ phần a $=> n_{HCl}=1,8 (mol)=>m_{HCl}=1,8.(1+35,5)=65,7(g)$$=>C $ %$=\frac{m_{HCl}}{m_{dd HCl}}.100$%$=36$
(%
)$<=>m_{dd HCl}=182,5(g)$Mà : $m=D.V=>V_{dd HCl}=182,5/1,18\approx 154,66(ml)$