Q: A
according to this formula, we can get the mole fraction of water (n):
P(solu) = n Pv(water)
when we have Pv(solu) = 22.8 and Pv(water) = 23.8 so by substitution:
22.8 = n * 23.8
n= 0.958
- we need to get the moles of glucose:
moles of water = 500 g(mass weight) / 18 (molar weight)= 27.7 mol
n = moles of water / ( moles of water + moles of glucose)
0.958 = 27.7 / ( 27.7+ moles of glucose)
0.958 moles of glucose + 26.5 = 27.7
0.968 moles of glucose = 1.2
moles of glucose = 1.253 mol
∴ the mass of glucose = no.of glucose moles x molar mass
= 1.253 x 180 = 225.5 g
Q: B
here we also need to get n (mole fraction of water )by using this formula:
Pv(solu) = n Pv(water)
when we have Pv(solu)=132 & Pv(water)=150 so, by substition:
132= n * 150
n = 0.88
so, mole fraction of solution = 1 - 0.88 = 0.12
and we can get after that the moles of water = (mass weight / molar mass)
- no.moles of water = 85 g / 18 g/mol = 4.7 moles
- total moles in solution = moles of water / moles fraction of water
= 4.7 / 0.88 = 5.34 moles
∴ moles of the solution = total moles in solu - moles of water
= 5.34 - 4.7 = 0.64 moles solute
∴ the molar mass of the solute = mass weight of solute / no.of moles of solute
= 53.8 / 0.64 = 84 g/mole
Q: C
moles of urea (NH2)2 CO = mass weight / molar mass
= 4.49 g / 60 g /mol
= 0.07 mol
moles of methanol = mass weight / molar mass
= 39.9 g / 32 g/mol = 1.25 mol
moles fraction of methanol = moles of methanol / (moles of methanol + moles of urea )
moles fraction of methanol = 1.25 / ( 1.25+0.07) = 0.95
by substitution in Pv formula we will be able to get the vapour pressure of the solu :
Pv(solu) = n P°v
Pv(solu) = 0.95 * 89 mm Hg
∴Pv(solu) = 84.55 mmHg
Box C will have the greatest density.
All boxes have the same volume.
Explanation:
We calculate the density using the following formula:
density = mass / volume
density of Box A = 10 g / 20 cm³ = 0.5 g/cm³
density of Box B = 30 g / 20 cm³ = 1.5 g/cm³
density of Box C = 170 g / 20 cm³ = 8.5 g/cm³
Box C will have the greatest density.
All boxes have the same volume.
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Answer:
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Explanation:
<u>1. Balanced molecular equation</u>
<u>2. Mole ratio</u>
<u>3. Moles of HNO₃</u>
- Number of moles = Molarity × Volume in liters
- n = 0.600M × 0.0100 liter = 0.00600 mol HNO₃
<u>4. Moles Ba(OH)₂</u>
- n = 0.700M × 0.0310 liter = 0.0217 mol
<u>5. Limiting reactant</u>
Actual ratio:
Since the ratio of the moles of HNO₃ available to the moles of Ba(OH)₂ available is less than the theoretical mole ratio, HNO₃ is the limiting reactant.
Thus, 0.006 moles of HNO₃ will react completely with 0.003 moles of Ba(OH)₂ and 0.0217 - 0.003 = 0.0187 moles will be left over.
<u>6. Final molarity of Ba(OH)₂</u>
- Molarity = number of moles / volume in liters
- Molarity = 0.0187 mol / (0.0100 + 0.0031) liter = 0.456M
