Đáp án:

Giải thích các bước giải:

 bài 4 

+) 1 mol KCl trong 750 ml dung dịch

$CM_{dd KCl}$ = $\frac{n}{V}$  = $\frac{1}{0,75}$ = 1,33(M)

+) 400g CuSO4 trong 4 lít dung dịch

$n_{CuSO4}$= $\frac{m}{M}$  = $\frac{400}{160}$=2,5 (mol)

⇒ $CM_{dd CuSO4}$ = $\frac{n}{V}$  = $\frac{2,5}{4}$ = 0,625(M)

+) 32 g khí SO2

$n_{SO2}$= $\frac{m}{M}$  = $\frac{32}{64}$=0,5 (mol)

⇒$V_{SO2}$=n . 22,4  =0,5 . 22,4=11,2(l)

vì $V_{SO2}$ ≈ $V_{ddSO2}$

⇒$CM_{dd SO2}$ = $\frac{n}{V}$  = $\frac{0,5}{11,2}$ = 0,04(M)

+) 0,5 mol MgCl2 trong 1,5 lít dd

$CM_{dd MgCl2}$ = $\frac{n}{V}$  = $\frac{0,5}{1,5}$ = 0,33(M)

+) 16g khí O2

$n_{O2}$= $\frac{m}{M}$  = $\frac{16}{36}$=0,44 (mol)

⇒$V_{O2}$=n . 22,4  =0,44 . 22,4=9,86(l)

vì $V_{O2}$ ≈ $V_{ddO2}$

⇒$CM_{dd O2}$ = $\frac{n}{V}$  = $\frac{0,44}{9,86}$ = 0,04(M)

+) 0,06 mol Na2CO3 trong 1500 ml dd

$CM_{dd Na2CO3}$ = $\frac{n}{V}$  = $\frac{0,06}{1,5}$ = 0,04(M)

a. 750ml=0,75 l
$C_{M_{KCl}}=\dfrac{n}{V}=\dfrac{1}{0,75}≈1,33(M)$
b. 
$C_{M_{MgCl_2}}=\dfrac{n}{V}=\dfrac{0,5}{1,5}≈0,33(M)$

c.
$n_{CuSO_4}=\dfrac{400}{160}=2,5(mol)$
$C_{M_{CuSO_4}}=\dfrac{n}{V}=\dfrac{2,5}{4}=0,625(M)$

d.
$n_{O_2}=\dfrac{16}{32}=0,5(mol)$
$V_{O_2}=Vdd_{O_2}=0,5.22,4=11,2(l)$
$C_{M_{O_2}}=\dfrac{n}{V}=\dfrac{0,5}{11,2}≈0,05(M)$
e.
$n_{SO_2}=\dfrac{32}{64}=0,5(mol)$
$V_{SO_2}=Vdd_{SO_2}=0,5.22,4=11,2(l)$
$C_{M_{SO_2}}=\dfrac{n}{V}=\dfrac{0,5}{11,2}≈0,05(M)$
f. $1500ml=1,5l$

$C_{M_{Na_2CO_3}}=\dfrac{n}{V}=\dfrac{0,06}{1,5}=0,04(M)$