This dilution problem uses the equation
M
a
V
a
=
M
b
V
b
M
a
= 6.77M - the initial molarity (concentration)
V
a
= 15.00 mL - the initial volume
M
b
= 1.50 M - the desired molarity (concentration)
V
b
= (15.00 + x mL) - the volume of the desired solution
(6.77 M) (15.00 mL) = (1.50 M)(15.00 mL + x )
101.55 M mL= 22.5 M mL + 1.50x M
101.55 M mL - 22.5 M mL = 1.50x M
79.05 M mL = 1.50 M
79.05 M mL / 1.50 M = x
52.7 mL = x
59.7 mL needs to be added to the original 15.00 mL solution in order to dilute it from 6.77 M to 1.50 M.
I hope this was helpful.
Hey there!:
Molar mass MgCl2 = 95.2110 g/mol
So:
1 mole MgCl2 -------------- 95.2110 g
moles MgCl2 ---------------- 319 g
moles MgCl2 = 319 * 1 / 95.2110
moles MgCl2 = 319 / 95.2110
=> 3.350 moles of MgCl2
Hope that helps!
Ernest Shackleton's South! primarily uses the writing structure of "problem and solution".
Answer:
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Answer:
Mass of solute = 0.0036 g
Explanation:
Given data:
Concentration of Cl⁻ = 15.0 ppm
Volume of water = 240 mL
Mass of Cl⁻ present = ?
Solution:
1 mL = 1 g
240 mL = 240 g
Formula:
ppm = mass of solute / mass of sample ×1,000,000
by putting values,
15.0 ppm = (mass of solute / 240 g) ×1,000,000
Mass of solute = 15.0 ppm × 240 g / 1,000,000
Mass of solute = 0.0036 g