Answer:
The number of solute particles increases, and the boiling point increases.
Explanation:
- It is known from colligative properties that adding solute to the solvent will cause elevation of boiling point.
- Elevation of boiling point (ΔTb) can be expressed as:
<em>ΔTb = Kb.m,</em>
where, Kb molal boiling point elevation constant.
m is the molal concentration of solute.
- Adding more sodium chloride to the solution:
will increase the number of solute particles and also will increase the molal concentration of NaCl solute.
<em>∵ ΔTb ∝ m.</em>
- So, the boiling point increases.
- Thus, the right choice is:
<em>The number of solute particles increases,</em>
<em></em>
<span>60. 4Li(s) + O2(g) >>>>>>>>>>> 2Li2O(s)
moles O2 required = (84 grams Li) * (1 mol Li / 6.941 grams Li) * (1 mol O2 / 4 mol Li) = 3.026 moles O2 required.
PV = nRT, P = 1 atm, V = ?, n = 3.026 moles, R = 0.0821 L*atm/mol*K, T = 273 K
(1 atm)*(V) = (3.026 moles O2) * (0.0821 L*atm/mol*K) * (273 K)
V = 67.82 Liters is the answer.
hope this helps</span>
Hi there,
I believe that your answer to this will be 234 g NaCL
this is because i made a reaction equation:
The reaction equation:
2Na + Cl₂ → 2NaCl
Molar ratio of Na to NaCl = 2 : 2 = 1
Moles of Na = 92/23
= 4
Moles of NaCl = 4
Mass NaCl = moles x Mr
= 234 grams
Hope this is correct :)
Have a great day