The number of moles and number of ions of each type in 88 mL of 1.75 M magnesium chloride is to be calculated. Concept introduction: Molarity ( M ) is one of the concentration terms that determine the number of moles of solute present in per liter of solution. Unit of molarity is mol/L . The expression to calculate the moles of the compound when molarity of solution and volume of solution are given is as follows: Moles of compound ( mol ) = [ volume of solution ( L ) ( molarity of solution ( mol ) 1L of solution ) ] The expression to calculate the moles of ions is as follows: moles of ion of compound ( mol ) = [ ( moles of compound ( mol ) ) ( total moles of ion ( mol ) 1 mole of compound ) ] The expression to calculate the number of ions is as follows: number of ions = ( moles of ions ( mol ) ) ( 6 .022 × 10 23 ions 1 mole of ions )
The number of moles and number of ions of each type in 88 mL of 1.75 M magnesium chloride is to be calculated. Concept introduction: Molarity ( M ) is one of the concentration terms that determine the number of moles of solute present in per liter of solution. Unit of molarity is mol/L . The expression to calculate the moles of the compound when molarity of solution and volume of solution are given is as follows: Moles of compound ( mol ) = [ volume of solution ( L ) ( molarity of solution ( mol ) 1L of solution ) ] The expression to calculate the moles of ions is as follows: moles of ion of compound ( mol ) = [ ( moles of compound ( mol ) ) ( total moles of ion ( mol ) 1 mole of compound ) ] The expression to calculate the number of ions is as follows: number of ions = ( moles of ions ( mol ) ) ( 6 .022 × 10 23 ions 1 mole of ions )
The number of moles and number of ions of each type in 88 mL of 1.75M magnesium chloride is to be calculated.
Concept introduction:
Molarity (M) is one of the concentration terms that determine the number of moles of solute present in per liter of solution. Unit of molarity is mol/L.
The expression to calculate the moles of the compound when molarity of solution and volume of solution are given is as follows:
Moles of compound(mol)=[volume of solution(L)(molarityofsolution(mol)1L of solution)]
The expression to calculate the moles of ions is as follows:
moles ofion of compound(mol)=[(moles of compound(mol))(total moles of ion(mol)1mole of compound)]
The expression to calculate the number of ions is as follows:
numberof ions=(moles of ions(mol))(6.022×1023ions1mole of ions)
(b)
Interpretation Introduction
Interpretation:
The number of moles and number of ions of each type in 321 mL of a solution containing 0.22 g aluminium sulfate per liter is to be calculated.
Concept introduction:
Molarity (M) is one of the concentration terms that determine the number of moles of solute present in per liter of solution. Unit of molarity is mol/L.
The expression to calculate the moles of solute when given mass and molecular mass of compound are given is as follows:
Moles of compound(mol)=[given massof compound(g)(1moleof compound(mol)molecular mass of compound(g))]
The expression to calculate the moles of ions is as follows:
moles ofion of compound(mol)=[(moles of compound(mol))(total moles of ion(mol)1mole of compound)]
The expression to calculate the number of ions is as follows:
numberof ions=(moles of ions(mol))(6.022×1023ions1mole of ions)
(c)
Interpretation Introduction
Interpretation:
The number of moles and number of ions of each type in 1.65 mL of a solution containing 8.83×1021 formula units of cesium nitrate per liter is to be calculated.
Concept introduction:
A formula unit is used for the ionic compound to represent their empirical formula. The expression to calculate the moles of a compound when the volume of solution and formula unit of a compound is given is as follows:
moles of a compound(mol)=[(volume of solution(L))(given formula unit of compound(FU))(1 mole of compound6.022×1023FU)]
The expression to calculate the moles of ions is as follows:
moles ofion of compound(mol)=[(moles of compound(mol))(total moles of ion(mol)1mole of compound)]
The expression to calculate the number of ions is as follows:
numberof ions=(moles of ions(mol))(6.022×1023ions1mole of ions)
Complete combustion of a 0.6250 g sample of the unknown crystal with excess O2 produced 1.8546 g of CO2 and 0.5243 g of H2O. A separate analysis of a 0.8500 g sample of the blue crystal was found to produce 0.0465 g NH3. The molar mass of the substance was found to be about 310 g/mol. What is the molecular formula of the unknown crystal?
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