Answer:
Solution given:
1 mole of KCl
22.4l
1 mole of KCl74.55g
we have
0.14 mole of KCl74.55*0.14=10.347g
74.55g of KCl22.4l
10.347 g of KCl22.4/74.55*10.347=3.11litre
volume of each solution contains 0.14 mol of KCl contain 3.11litre.
Answer:
you did not show the options
Explanation:
I just got this question and it was lipids sorry if i’m wrong
Answer:
120.0 mL.
Explanation:
- As it is known that the no. of millimoles of a solution before dilution is equal to the no. of millimoles after dilution.
We suppose that the initial W% of methanol is 100.0 %
<em>∴ (W%V) before dilution = (W%V) after dilution.</em>
W% before dilution = 100.0 %, V before dilution = 18.0 mL.
W% after dilution = 15.0 %, V after dilution = ??? mL.
<em>∴ V after dilution = (W%V) before dilution/W% after dilution = </em>(100.0 %)(18.0 mL)/(15.0%)<em> = 120.0 mL.</em>
