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the number of p atoms in 1.0 g of ba₃(po₄)₂ is: a) 5.0 × 10⁻²² b) 2.0 ×…

Question

the number of p atoms in 1.0 g of ba₃(po₄)₂ is: a) 5.0 × 10⁻²² b) 2.0 × 10²¹ c) 5.0 × 10²⁰ d) 7.2 × 10²⁶ e) 1.0 × 10²¹

Explanation:

Step1: Calculate molar mass of \( Ba_3(PO_4)_2 \)

Molar mass of \( Ba \): \( 137.33 \, g/mol \), \( P \): \( 30.97 \, g/mol \), \( O \): \( 16.00 \, g/mol \)
Molar mass \( = 3\times137.33 + 2\times(30.97 + 4\times16.00) \)
\( = 412.0 + 2\times(30.97 + 64.00) \)
\( = 412.0 + 2\times94.97 \)
\( = 412.0 + 189.94 = 601.94 \, g/mol \approx 602 \, g/mol \)

Step2: Moles of \( Ba_3(PO_4)_2 \) in 1.0 g

Moles \( = \frac{mass}{molar\ mass} = \frac{1.0\ g}{602\ g/mol} \approx 1.66\times10^{-3}\ mol \)

Step3: Moles of P atoms

In \( Ba_3(PO_4)_2 \), 1 mole has 2 moles of P.
Moles of P \( = 2 \times 1.66\times10^{-3}\ mol \approx 3.32\times10^{-3}\ mol \)

Step4: Number of P atoms (using Avogadro's number \( 6.022\times10^{23}\ mol^{-1} \))

Number of atoms \( = moles \times N_A = 3.32\times10^{-3}\ mol \times 6.022\times10^{23}\ mol^{-1} \approx 2.0\times10^{21} \) (Wait, correction: Wait, earlier step2 moles of \( Ba_3(PO_4)_2 \) is \( 1/602 \approx 1.66\times10^{-3} \), moles of P: 2 1.66e-3 = 3.32e-3? No, wait 1 mole \( Ba_3(PO_4)_2 \) has 2 moles P. So moles P = 2 (1/602) ≈ 3.32e-3? No, 1/602 is ~1.66e-3, times 2 is ~3.32e-3? Wait no, 1/602 is ~0.00166, times 2 is ~0.00332. Then number of atoms: 0.00332 6.02e23 ≈ 2.0e21? Wait no, wait 0.00332 6e23 = 1.99e21 ≈ 2.0e21? Wait but let's recalculate:

Wait molar mass: \( Ba_3(PO_4)_2 \): 3137.33=412, 2P=61.94, 8*O=128. So total: 412+61.94+128=601.94, correct.

Moles of \( Ba_3(PO_4)_2 \): 1/602 ≈ 0.00166 mol.

Moles of P: 2 * 0.00166 ≈ 0.00332 mol.

Number of P atoms: 0.00332 6.02e23 = 0.003326e23 = 1.99e21 ≈ 2.0e21. Wait, but wait, maybe my initial molar mass was wrong? Wait no, let's check again. Wait, 3 Ba: 3137.33=412, 2 P: 230.97=61.94, 8 O: 8*16=128. Sum: 412+61.94=473.94+128=601.94, correct.

Wait, but maybe I made a mistake in moles of P. Wait, formula is \( Ba_3(PO_4)_2 \), so each formula unit has 2 P atoms. So moles of P = 2 moles of \( Ba_3(PO_4)_2 \). So moles of \( Ba_3(PO_4)_2 \) is 1/602 ≈ 0.00166 mol. Then moles of P is 20.00166 ≈ 0.00332 mol. Then number of P atoms: 0.00332 * 6.02e23 = 1.99e21 ≈ 2.0e21. So option b.

Wait, but let's recalculate with more precise numbers:

Molar mass: 3137.33 = 411.99, 230.97=61.94, 8*16=128. Sum: 411.99+61.94=473.93+128=601.93 g/mol.

Moles of \( Ba_3(PO_4)_2 \): 1.0 / 601.93 ≈ 0.001661 mol.

Moles of P: 2 * 0.001661 ≈ 0.003322 mol.

Number of P atoms: 0.003322 6.022e23 = 0.0033226.022e23 ≈ 2.00e21. So correct option is b.

Wait, but earlier when I thought maybe mistake, but no. So the answer is b.

Answer:

b) \( 2.0 \times 10^{21} \)