1.3 Liuoksen konsentraatio

20
a) 1,5mol/l
b) 3,7mmol/l

21
n=\frac{m}{M}
M=2\cdot14{,}01+8\cdot1{,}008+32{,}07+4\cdot16{,}00=72{,}314\ \frac{g}{mol}
n=0{,}041485...mol
c=\frac{n}{V}=\frac{0{,}0418...mol}{0{,}1l}=0{,}4148...\ \frac{mol}{l}\approx0{,}4\ \frac{mol}{l}
\left(NH_4\right)_2SO_4\rightarrow\ 2\ NH_4^++SO_4^{2-}
c\left(NH_4^+\right)=2\cdot0{,}4148...\ \frac{mol}{l}=0{,}8297...\ \frac{mol}{l}\approx0{,}8\ \frac{mol}{l}
c\left(SO_4^{2-}\right)=0{,}4148...\ \frac{mol}{l}\approx0{,}4\ \frac{mol}{l}
23
a)
c=\frac{n}{V}
V=\frac{n}{c}=\frac{0{,}10mol}{0{,}14\ \frac{mol}{l}}=0{,}7142857...l\approx714{,}29ml\approx710ml
b)
n=\frac{m}{M}
m=0{,}001g
M\left(NaCl\right)=22{,}99+35{,}45=58{,}44\ \frac{g}{mol}
n=0{,}001711156...mol
c=\frac{n}{V}\
V=\frac{n}{c}=\frac{0{,}001711156...mol}{0{,}14\ \frac{mol}{l}}=0{,}000122225...l\approx0{,}12ml
c)
M\left(Na^+\right)=22{,}99\ \frac{g}{mol}
m=nM=0{,}14\ mol\cdot22{,}99\ \frac{g}{mol}=3{,}2186...\ \frac{g}{l}\approx3{,}2\ \frac{g}{l}
d)
0{,}14\ \frac{mol}{l}\cdot0{,}50l=0{,}07mol
0{,}07mol\cdot N_A=4{,}2154...\cdot10^{22}


25
m=cV=0{,}004\ \frac{mol}{l}\cdot0{,}2l=0{,}0008mol=0{,}8mmol
M\left(K^+\right)=39{,}10\ \frac{g}{mol}
m=nM=0{,}0008mol\cdot39{,}10\ \frac{g}{mol}=0{,}03128g\approx31{,}28mg

työ 4, osa 1:
V=100ml
c=0{,}10\ \frac{mol}{l}
n=cV=0{,}1l\cdot0{,}10\ \frac{mol}{l}=0{,}01mol
n=\frac{m}{M}
M\left(C_{12}H_{22}O_{11}\right)=12\cdot12{,}01+22\cdot1{,}008+11\cdot16{,}00=342{,}296\ \frac{g}{mol}
m=nM=0{,}01mol\cdot342{,}296\ \frac{g}{mol}=3{,}422...g\approx3{,}4g
osa 2:
V_1=\frac{1}{5}V_2=\frac{1}{5}\cdot100ml=20ml

27
a)
1180g\cdot0{,}36=424{,}8g
M\left(HCl\right)=36{,}46\ \frac{g}{mol}
n=\frac{m}{M}=\frac{424{,}8g}{36{,}46\ \frac{g}{mol}}=11{,}651124...mol\approx12mol
c=12\ \frac{mol}{l}
b)
910g\cdot0{,}25=227{,}5g
M\left(NH_3\right)=17{,}03\ \frac{g}{mol}
n=\frac{m}{M}=\frac{227{,}5g}{17{,}03\ \frac{g}{mol}}=13{,}358778...mol\approx13mol
c=13\ \frac{mol}{l}