Answer:-
0.229 L
Explanation:-
Molar mass of AgBr = 107.87 x 1 + 79.9 x 1
=187.77 grams mol-1
Mass of AgBr = 150 grams
Number of moles of AgBr = 150 grams / 187.77 gram mol-1
= 0.8 mol
The balanced chemical equation is
NaBr (aq) + AgNO3 (aq)--> AgBr(s) + NaNO3(aq)
From the equation we can see that
1 mol of AgBr is produced from 1 mol of AgNO3.
∴ 0.8 mol of AgBr is produced from 1 x 0.8 / 1 = 0.8 mol of AgNO3.
Strength of AgNO3 = 3.5 M
Volume of AgNO3 required = Number of moles / strength
= 0.8 moles / 3.5
=0.229 L
Answer : The mass of sodium bromide added should be, 18.3 grams.
Explanation :
Molality : It is defined as the number of moles of solute present in kilograms of solvent.
Formula used :
Solute is, NaBr and solvent is, water.
Given:
Molality of NaBr = 0.565 mol/kg
Molar mass of NaBr = 103 g/mole
Mass of water = 315 g
Now put all the given values in the above formula, we get:
Thus, the mass of sodium bromide added should be, 18.3 grams.
The total volume of water that would be removed will be 75 mL
<h3>Dilution equation</h3>
Using the dilution equation:
M1V1 = M2V2
In this case, M1 = 500 mL, V1 = 10.20 M, M2 = 12 M
Substitute:
V2 = 500 x 10.20/12
= 425 mL
The final volume in order to arrive at 12 M HNO3 would be 425 mL from the initial 500 mL. Thus, the total amount of water that will be removed by evaporation can be calculated as:
500 - 425 = 75 mL
More on dilution can be found here: brainly.com/question/7208939
