Question

what mass of magnesium phosphate will contain the same number of molecules as present in 4.1 gram of calcium nitrate given the atomic masses are.​
phosphate=31u
calcium=40u
magnesium=24u
oxygen=16u
nitrogen=14u

Answer 1

Answer:

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Explanation:

262.86 is the mass of magnesium phosphate

Answer 2

Answer:

The required mass of magnesium phosphate is 6.55 g.

Step-by-step-explanation:

We have given that,

• Mass of calcium nitrate ( Ca(NO₃)₂ ) = 4.1 g
• Atomic mass of phosphorus ( P ) = 31 u
• Atomic mass of calcium ( Ca ) = 40 u
• Atomic mass of magnesium ( Mg ) = 24 u
• Atomic mass of oxygen ( O ) = 16 u
• Atomic mass of nitrogen ( N ) = 14 u

Now, we know that,

Molecular formula of calcium nitrate = Ca(NO₃)₂

Molar mass of Ca(NO₃)₂ = Atomic mass of Ca + Atomic mass of N * 6 + Atomic mass of O * 6

⇒ Molar mass of Ca(NO₃)₂ = 40 + 14 * 2 + 16 * 6

⇒ Molar mass of Ca(NO₃)₂ = 40 + 28 + 96

⇒ Molar mass of Ca(NO₃)₂ = 40 + 24 + 96 + 4

⇒ Molar mass of Ca(NO₃)₂ = 40 + 120 + 4

⇒ Molar mass of Ca(NO₃)₂ = 160 + 4

⇒ Molar mass of Ca(NO₃)₂ = 164 g/mol

Now, we know that,

Number of moles = Given mass / Molar mass

⇒ No. of moles of Ca(NO₃)₂ = Given mass of Ca(NO₃)₂ / Molar mass of Ca(NO₃)₂

⇒ No. of moles of Ca(NO₃)₂ = 4.1 g / 164 g/mol

⇒ No. of moles of Ca(NO₃)₂ = ( 4.1 / 164 ) mol

⇒ No. of moles of Ca(NO₃)₂ = ( 410 / 1640 ) mol

⇒ No. of moles of Ca(NO₃)₂ = ( 82 / 3280 ) mol

⇒ No. of moles of Ca(NO₃)₂ = [ 82 / ( 82 * 40 ) ] mol

⇒ No. of moles of Ca(NO₃)₂ = ( 82 ÷ 82 * 1 / 40 ) mol

⇒ No. of moles of Ca(NO₃)₂ = 1 / 40 mol

⇒ No. of moles of Ca(NO₃)₂ = 0.025 moles

We have to find the mass of magnesium phosphate for which the number of molecules are same as present in 4.1 g of calcium nitrate.

Now, we know that,

Number of molecules = No. of moles * Avogadro's number

No. of molecules of Ca(NO₃)₂ = No. of molecules of Mg₃(PO₄)₂

⇒ No. of moles of Ca(NO₃)₂ * Nᴀ = No. of moles of Mg₃(PO₄)₂ * Nᴀ

⇒ No. of moles of Ca(NO₃)₂ = No. of moles of Mg₃(PO4)₂

⇒ No. of moles of Mg₃(PO₄)₂ = 0.025 moles

Now,

Molar mass of Mg₃(PO₄)₂ = Atomic mass of Mg * 3 + Atomic mass of P * 2 + Atomic mass of O * 8

⇒ Molar mass of Mg₃(PO₄)₂ = 24 * 3 + 31 * 2 + 16 * 8

⇒ Molar mass of Mg₃(PO₄)₂ = 72 + 62 + 128

⇒ Molar mass of Mg₃(PO₄)₂ = 72 + 190

⇒ Molar mass of Mg₃(PO₄)₂ = 262 g/mol

We know that,

Number of moles = Given mass / Molar mass

⇒ No. of moles of Mg₃(PO₄)₂ = Required mass / Molar mass of Mg₃(PO₄)₂

⇒ 0.025 mol = Required mass / 262 g/mol

⇒ Required mass = 0.025 * 262 mol * g/mol

⇒ Required mass = 6.55 g

⇒ Required mass of Mg₃(PO₄)₂ = 6.55 g

∴ The required mass of magnesium phosphate is 6.55 g.