Use M1V1 = M2V2 to solve
3(V1) = 2.8 * 1.6
3(V1) = 4.48
V1 = 1.493 L of stock solution
Mass of PH3= 6.086 g
<h3>Further explanation</h3>
Given
6.0 L of H2
Required
mass of PH3
Solution
Reaction
P4 + 6H2 → 4PH3
Assumed at STP ( 1 mol gas=22.4 L)
Mol of H2 for 6 L :
= 6 : 22.4 L
= 0.268
From the equation, mol PH3 :
= 4/6 x moles H2
= 4/6 x 0.268
= 0.179
Mass PH3 :
= 0.179 x 33,99758 g/mol
= 6.086 g
To solve this questions you first need to find the number of moles of barium phosphate you have. The molar mass of barium phosphate is 601.93g/mol.
24.4/601.83 = 0.0402 moles barium phosphate
Then you need to use avagadro’s number, 6.022 x 10^23, which is the number of molecules or formula units in a mole.
6.022 x 10^23 * 0.0402 = 2.42 x 10^22 formula units
<u>Answer:</u> The freezing point of solution is -0.454°C
<u>Explanation:</u>
Depression in freezing point is defined as the difference in the freezing point of pure solution and freezing point of solution.
The equation used to calculate depression in freezing point follows:
To calculate the depression in freezing point, we use the equation:
Or,
where,
Freezing point of pure solution = 0°C
i = Vant hoff factor = 2
= molal freezing point elevation constant = 1.86°C/m
= Given mass of solute (KCl) = 5.0 g
= Molar mass of solute (KCl) = 74.55 g/mol
= Mass of solvent (water) = 550.0 g
Putting values in above equation, we get:
Hence, the freezing point of solution is -0.454°C
