1 pm equals 10^{-10} cm. Therefore, 230 pm can be converted to 2.3 × 10^{-8} cm. Atoms are spherical in shape, and the volume of a sphere can be calculated using the formula 4/3πr³. Consequently, the atom's volume is calculated as 4/3π(2.3 × 10^{-8})³. Evaluating this gives 4/3 × (3.142 × 12.167 × 10^{-24}), which simplifies to approximately 5.096 × 10^{-23} cm³. Since 1 m³ is equal to 1,000,000 cm³, we find that the volume of the atom is 5.096 × 10^{-29} m³.
The correct answer is Option A.
The calculation goes as follows:
Number of millimoles of Na3PO4 = 1 × 100 = 100
Number of millimoles of AgNO3 = 1 × 100 = 100
Dissociating 1 mole of Na3PO4 yields 3 moles of sodium ions and 1 mole of phosphate ions, whereas 1 mole of AgNO3 releases 1 mole of Ag+ and 1 mole of NO3-.
The Ag+ ion concentration becomes negligible since it forms a precipitate with the phosphate ion, indicating that the concentration of phosphate ions is also low.
With 100 millimoles of Na3PO4, we get 300 millimoles of Na+ and 100 millimoles of PO43-, and with 100 millimoles of AgNO3 we have 100 millimoles of Ag+ and 100 millimoles of NO3-.
Thus, the order of increasing concentration is: PO43- < NO3- < Na+.
The question is incomplete,the complete question:
Determine the molality of a 10.0% (by weight) solution of hydrochloric acid in water:
a) 0.274 m
b) 2.74 m
c) 3.05 m
d) 4.33 m
e) the solution's density is necessary for calculations
Answer:
The molality for a 10.0% (by weight) hydrochloric acid solution is 3.05 mol/kg.
Explanation:
The solution is a 10.0% (by weight) hydrochloric acid mix.
This means there are 10 grams of HCl in 100 grams of the solution.
Amount of HCl = 10 g
Total mass of solution = 100 g
Total mass of solution = Mass of solute + Mass of solvent
Mass of solvent (water) = 100 g - 10 g = 90 g
Calculate moles of HCl = 
Mass of water converted to kilograms = 0.090 kg
Molality = 
<strongTherefore, the molality of a 10.0% (by weight) hydrochloric acid solution is 3.05 mol/kg.
