What volume (in L) of a 4.13 M lithium nitrate $L i N O_{3}$ $LiNO_3$ solution would be needed to make 195 mL of a 1.05 M solution by dilution?

Question

What volume (in L) of a 4.13 M lithium nitrate $L i N O_{3}$ $LiNO_3$ solution would be needed to make 195 mL of a 1.05 M solution by dilution?

1 Answer

Impact of this question

5484 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-05-17T07:33:30

$C_{1} V_{1} = C_{2} V_{2}$ $C_1V_1=C_2V_2$

$V_{2} ≅ \cdot 50 \cdot m L$ $V_2~=*50*mL$

Explanation:

$Concentration$ $"Concentration"$ $=$ $=$ $\frac{Moles of solute}{Volume of solution}$ $"Moles of solute"/"Volume of solution"$

Thus $Moles$ $"Moles"$ $=$ $=$ $Volume \times concentration$ $"Volume "xx" concentration"$

So $C_{1} V_{1} = C_{2} V_{2}$ $C_1V_1=C_2V_2$ , where $C = Concentration; V=Volume$ $C="Concentration; V=Volume"$

And $V_{1} = \frac{C_{2} V_{2}}{C_{1}}$ $V_1=(C_2V_2)/C_1$ $=$ $=$ $\frac{1.05 \cdot m o l \cdot L^{- 1} \times 0.195 \cdot L}{4.13 \cdot m o l \cdot L^{- 1}}$ $(1.05*mol*L^-1xx0.195*L)/(4.13*mol*L^-1)$ $=$ $=$ $? ? L$ $??L$

Science

Math

Humanities

... and beyond

What volume (in L) of a 4.13 M lithium nitrate $L i N O_{3}$ $LiNO_3$ solution would be needed to make 195 mL of a 1.05 M solution by dilution?

1 Answer

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

What volume (in L) of a 4.13 M lithium nitrate LiNO_3LiNO3LiNO_3 solution would be needed to make 195 mL of a 1.05 M solution by dilution?

1 Answer

Explanation:

Related questions

Impact of this question

What volume (in L) of a 4.13 M lithium nitrate $L i N O_{3}$ $LiNO_3$ solution would be needed to make 195 mL of a 1.05 M solution by dilution?