10.5g of Cu(NO3)2 are dissolved in 500g of water. Determine the expected boiling point of the solution. (Please)?

Molar mass of Cu(NO₃)₂ = (63.5 + 14.0×2 + 16.0×6) g/mol = 187.5 g/mol
No. of moles of Cu(NO₃)₂ = (10.5 g) / (187.5 g/mol) = 0.0560 mol

Each mole of Cu(NO₃)₂ dissociates in aqueous solution to give 3 moles of ions (1 mole of Cu²⁺ and 2 moles of NO₃⁻).
van't Hoff factor, i = 3
molality of Cu(NO₃)₂, m = (0.0560 mol) / (500/1000 kg) = 0.112 mol/kg
Ebullioscopic constant, Kb = 0.512 °C kg/mol

Boiling point elevation, ΔTb = i m Kb = 3 × (0.112 mol/kg) × (0.512 °C kg/mol) = 0.17 °C

Normal boiling point of water = 100 °C
Expected boiling point of the solution = (100 + 0.17) °C = 100.17 °C

Boiling point elevation.....

ΔT = (i)(Kb)(c) ............. where i is the van't Hoff factor, which for Cu(NO3)2 will be essentially "3"; Kb is the boiling point elevation constant for water which is 0.512 C/m; c is the concentration if molality (moles of solute per kilogram of water).

ΔT = 3 x 0.512C/m x (10.5g x (1 mol / 187.5g) / 0.500 kg)
ΔT = 0.172C

This is if we assume that "500g of water" is expressed to three significant digits. You could compute the molality separately instead of folding it into the chain calculation.
molality = moles / kilograms
molality = 10.5g x (1 mol / 187.5g) / 0.500 kg = 0.112 m

ΔT = (i)(Kb)(c) = 3 x 0.512C/m x 0.112m =0.172C

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Either way, the boiling point of the copper(II) nitrate solution will be 100.172C.
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It is important to note that the integer value of "i" is an approximation and is likely higher than a more realistic value would be. On the surface we assume that Cu(NO3)2 will ionize to form three particles for each formula unit of Cu(NO3)2. In reality, that value will be less than 3. That is due to the incomplete ionization of the salt, the hydrolysis of Cu^2+ (forming [CuOH]^+) and to ion-pairing. Nor is the value of "i" fixed at some particular number. Instead, it will vary, depending on the concentration of the salt. The more dilute the salt, the more the value of "i" will approach the ideal number. Therefore, when you pick a value of "3" for the value of "i" you are making an approximation, and it is unlikely that the boiling point will be as high as indicated in the calculation above.