It would be C
2 kg x 1000 g/kg x 1mol/18.02 x 6.03 kj/mol = 669kj
Answer:
2.2 °C/m
Explanation:
It seems the question is incomplete. However, this problem has been found in a web search, with values as follow:
" A certain substance X melts at a temperature of -9.9 °C. But if a 350 g sample of X is prepared with 31.8 g of urea (CH₄N₂O) dissolved in it, the sample is found to have a melting point of -13.2°C instead. Calculate the molal freezing point depression constant of X. Round your answer to 2 significant digits. "
So we use the formula for <em>freezing point depression</em>:
In this case, ΔTf = 13.2 - 9.9 = 3.3°C
m is the molality (moles solute/kg solvent)
- 350 g X ⇒ 350/1000 = 0.35 kg X
- 31.8 g Urea ÷ 60 g/mol = 0.53 mol Urea
Molality = 0.53 / 0.35 = 1.51 m
So now we have all the required data to <u>solve for Kf</u>:
Answer:
CuSO4 + 2NaOH → Cu(OH)2 + Na2SO4
Explanation:
Answer:
0.5M is the answer.
Explanation:
1M solution is the solution containing 1mole solute dissolved per litre of solution.
Using unitary method,
1000cc gives 1M.
1cc gives 1/1000M.
500 cc gives 500/1000M=0.5M
