Answer:
55.75g
Explanation:
From
m/M = CV
Where
m= required mass of solute
M= molar mass of solute
C= concentration of solution
V= volume of solution=675ml
Molar mass of solute= 3(23) + 31 + 4(16)= 69+31+64=164gmol-1
Number of moles of sodium ions present= 1.5× 675/1000= 1.01 moles
Since 1 mole of Na3PO4 contains 3 moles of Na+
It implies that 1.01/3 moles of Na3PO4 are present in solution= 0.34moles
mass of Na3PO4= number of moles × molar mass= 0.34 × 164 =55.75g
The answer to your question is true.
<u>Answer:</u>
<em>The final volume must be 400 mL.</em>
<em></em>
<u>Explanation:</u>
Let us assume the Initial volume as 100ml
Using dilution factor formula
So,
Thus, the final volume must be 400 mL.
Answer:
D. 20.6g
Explanation:
Molarity is defined as ratio between moles of solute (sodium bromide, NaBr) per liter of water.
As density of water is 1g/mL; volume of 400.0g of water is 400.0mL = 0.4000L.
That means 0.400L of 0.500M solution contains:
0.400L × (0.500mol / 1L) = <em>0.200moles of sodium bromide</em>.
In mass (NaBr = 102.9g/mol):
0.200mol NaBr × (102.9g/mol) = 20.6g of NaBr
Right answer is:
<em>D. 20.6g</em>
