Đáp án + Giải thích các bước giải:

$a\big)$

Đổi $3l=3000ml$

$\to m_{dd\,NaOH}=V_{dd\,NaOH}.d=3000\times 1,115=3345(g)$

$\to m_{CT\,NaOH}=C\%.m_{dd\,NaOH}=3345\times 10\%=334,5(g)$

$\to m_{H_2O}=m_{dd\,NaOH}-m_{CT\,NaOH}=3345-334,5=3010,5(g)$

Ta có $n_{NaOH}=\dfrac{334,5}{40}=8,3625(g)$

$\to C_{M_{NaOH}}=\dfrac{8,3625}{3}=2,7875M$

$b\big)$

Ta có $m_{CT\,AgNO_3}=C\%.m_{dd\,AgNO_3}=700\times 20\%=140(g)$

$\to m_{H_2O}=m_{dd\,AgNO_3}-m_{CT\,AgNO_3}=700-140=560g)$

`a)`

`V={m_(dd)}/{D_(dd)}\Rightarrow m_(dd)=D_(dd).V_(dd)=3000.1,115=3345g`

`\Rightarrow m_(ct)={mdd.C%}/{100%}={3345.10%}/{100%}=334,5g`

`\Rightarrow m_(NaOH)=m_(ct)=334,5g`

`b)`

`m_{AgNO_3}=C\%.m_{AgNO_3}=700.20%=140g`

`-> m_{H_2O}=m_{AgNO_3}-m_{AgNO_3}=700-140=560g`