Đáp án:

 

Giải thích các bước giải:

Bài 1 :

$m_{dd} = 20 + 180 = 200 (gam)$

$⇒ C\%_{NaCl} = \frac{20}{200}.100\% = 10\%$

Bài 2 :

$m_{HCl} = 300.10\% = 30(gam)$

Bài 3 :

$m_{dd(KOH)} = \frac{m_{KOH}}{C\%} = \frac{15}{5\%} = 300(gam)$

Bài 4 :

$m_{NaOH} = 200.3\% + 200.7\% = 20(gam)$
$m_{dd} = 200 + 200 = 400(gam)$
$⇒ C\%_{NaOH} = \frac{20}{400}.100\% = 5\%$

Bài 5 :

$n_{H2SO4} = \frac{9,8}{98} = 0,1(mol)$

$⇒ C_M H_2SO_4 = \frac{0,1}{4} = 0,025M$

Bài 6 :

$n_{MgCl2} = 0,2.2 = 0,4(mol)$
$⇒ m_{MgCl2} = 0,4.95 = 38(gam)$
Bài 7 :

$n_{HCl} = \frac{7,3}{36,5} = 0,2(mol)$
$⇒ V_{HCl(1M)} = \frac{0,2}{1} = 0,2(lít)$

Bài 8 :

có : $n_{KCl} = 0,2.1 + 0,3.2 = 0,8(mol)$
Sau khi trộn , $V_{dd} = 200 + 300 = 500(ml) = 0,5(lít)$
$⇒ C_M KCl = \frac{0,8}{0,5} = 1,6M$

 

`1.`

`C\%={20}/{20+180}.100\%=10\%`

`2.`

`m_{HCl}=300.10\%=30(g)`

`3.`

`m_{KOH}=15.5\%=0,75(g)`

`4.`

`C\%_A={200.3\%+200.7\%}/{200+200}.100\%=5\%`

`5.`

`n_{H_2SO_4}={9,8}/{98}=0,1(mol)`

`C_{M\ H_2SO_4}={0,1}/4=0,025M`

`6.`

`n_{MgCl_2}=0,2.2=0,4(mol)`

`m_{MgCl_2}=0,4.95=38(g)`

`7.`

`n_{HCl}={7,3}/{36,5}=0,2(mol)`

`V_{dd\ HCl}={0,2}/1=0,2(l)=200(ml)`

`8.`

`C_{M\ KCl\ sau}={0,2.1+0,3.2}/{0,2+0,3}=1,6M`