Answer: 1.997 M
Explanation:
molarity = moles of solute/liters of solution or
first we have to find our moles of solute (mol), which you can find by dividing the mass of solute by molar mass of solute
mass of solute: 92 g
molar mass of solute: 46.08 g/mol
let's plug it in:
next, we plug it into our original equation:
Mass percentage of sodium chloride(NaCl) in ocean waters = 3.5 %
That means 3.5 g sodium chloride(NaCl) is present for every 100 g of ocean water.
The given mass of sodium chloride(NaCl) is 45.8 g
Calculating the mass of ocean waters that would contain 45.8 g sodium chloride(NaCl):
= 1309 g ocean water
Therefore, 45.8 g sodium chloride is present in 1309 g ocean water.
Answer:
a) molarity of CCl3F = 1.12 × 10^-11 mol/dm³
Molarity of CCl2F2 = 2.20 × 10^-11 mol/dm³
B) molarity of CCL3F = 7.96 × 10 ^-13 mol/dm³
Molarity of CCl2F2 = 1.55 × 10^-12 mol/dm³
Explanation:
Using the ideal gas equation:
PV = nRT
Further explanations are found in the attachment below.
