The grams of glucose are needed to prepare 400g of a 2.00%(m/m) glucose solution g is calculated as below
=% m/m =mass of the solute/mass of the solution x100
let mass of solute be represented by y
mass of solution = 400 g
% (m/m) = 2% = 2/100
grams of glucose is therefore =2/100 = y/400
by cross multiplication
100y = 800
divide both side by 100
y= 8.0 grams
Answer:
a. 5.9 × 10⁻³ M/s
b. 0.012 M/s
Explanation:
Let's consider the following reaction.
2 N₂O(g) → 2 N₂(g) + O₂(g)
a.
Time (t): 12.0 s
Δn(O₂): 1.7 × 10⁻² mol
Volume (V): 0.240 L
We can find the average rate of the reaction over this time interval using the following expression.
r = Δn(O₂) / V × t
r = 1.7 × 10⁻² mol / 0.240 L × 12.0 s
r = 5.9 × 10⁻³ M/s
b. The molar ratio of N₂O to O₂ is 2:1. The rate of change of N₂O is:
5.9 × 10⁻³ mol O₂/L.s × (2 mol N₂O/1 mol O₂) = 0.012 M/s
Boiling point of a compound is determined by the strength of intermolecular forces of attraction between the molecules present in it. Stronger the intermolecular forces of attraction, higher will be the boiling point.
Ionic compounds show ion-ion interactions which are the strongest among all. Ion-dipole interactions are shown when ionic solutes are dissolved in polar solvents. Hydrogen bonding is also a relatively stronger force that is present between H atom and an electronegative atom like F, O and N(
) . All polar molecules show dipole-dipole interaction (
and 
). Dispersion forces are the weakest intermolecular forces due to momentary dipoles between electron clouds and nucleus.
Among the given compounds, 
has dispersion forces as the major intermolecular forces of attraction. So they they exhibit the weakest IMF, hence have the lowest boiling point.
