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Hint: To solve this question, follow these steps: First, identify the molecular formula of the given compound. Then calculate the molecular weight of sodium hydroxide. Then, to find the number of moles of the given compound, divide the mass of the compound in the sample with the molecular mass of the compound.
Formula used: Number of moles of NaOH \[ = \dfrac{{Weight\,\,of\,\,the\,\,given\,\,sample}}{{molecular\,\,weight\,\,of\,\,NaOH}}\]
Complete Step-by-Step Answer:
Before we move forward with the solution of this question, let us discuss some important concepts.
Mole: a mole is a physical quantity which represents the amount of mass of the substance required to have a collective of \[6.022 \times {10^{23}}\] atoms of the given substance. Mole is a widely used unit for calculating the amount of matter of a substance. One mole of any substance weighs about the same as the molecular mass of that substance.
Now moving back to the question, we have been given a sample of 160 grams of Sodium hydroxide. The molecular formula of sodium hydroxide is NaOH. We know that the atomic weight of hydrogen is 1 amu, atomic weight of sodium is 23 amu and the atomic weight of oxygen is 16 amu. Hence the molecular weight of \[HN{O_3}\] can be calculated as:
Molecular weight of \[HN{O_3}\] = (no. of atoms of sodium) (atomic weight of sodium) +
(no. of atoms of hydrogen) (atomic weight of hydrogen) +
(no. of atoms of oxygen) (atomic weight of oxygen)
\[ = \left( 1 \right)\left( {23} \right) + \left( 1 \right)\left( 1 \right) + \left( 1 \right)\left( {16} \right)\]
= 40 \[g/mol\]
Now, to calculate the number of moles of NaOH present in the given sample, we use the formula:
Number of moles of NaOH \[ = \dfrac{{Weight\,\,of\,\,the\,\,given\,\,sample}}{{molecular\,\,weight\,\,of\,\,NaOH}}\] \[ = \dfrac{{160}}{{40}}\]
Number of moles of NaOH = 4 moles
Hence, the number of moles of sodium hydroxide contained in 160 g of it is 4 moles.
Note: Mole concept simplifies the mass relation among reactants and products such that we can base our calculation on the coefficients (numbers of molecules involved in the reaction). At the same time, mass or the quantity of substance is on lab scale in grams.
Formula used: Number of moles of NaOH \[ = \dfrac{{Weight\,\,of\,\,the\,\,given\,\,sample}}{{molecular\,\,weight\,\,of\,\,NaOH}}\]
Complete Step-by-Step Answer:
Before we move forward with the solution of this question, let us discuss some important concepts.
Mole: a mole is a physical quantity which represents the amount of mass of the substance required to have a collective of \[6.022 \times {10^{23}}\] atoms of the given substance. Mole is a widely used unit for calculating the amount of matter of a substance. One mole of any substance weighs about the same as the molecular mass of that substance.
Now moving back to the question, we have been given a sample of 160 grams of Sodium hydroxide. The molecular formula of sodium hydroxide is NaOH. We know that the atomic weight of hydrogen is 1 amu, atomic weight of sodium is 23 amu and the atomic weight of oxygen is 16 amu. Hence the molecular weight of \[HN{O_3}\] can be calculated as:
Molecular weight of \[HN{O_3}\] = (no. of atoms of sodium) (atomic weight of sodium) +
(no. of atoms of hydrogen) (atomic weight of hydrogen) +
(no. of atoms of oxygen) (atomic weight of oxygen)
\[ = \left( 1 \right)\left( {23} \right) + \left( 1 \right)\left( 1 \right) + \left( 1 \right)\left( {16} \right)\]
= 40 \[g/mol\]
Now, to calculate the number of moles of NaOH present in the given sample, we use the formula:
Number of moles of NaOH \[ = \dfrac{{Weight\,\,of\,\,the\,\,given\,\,sample}}{{molecular\,\,weight\,\,of\,\,NaOH}}\] \[ = \dfrac{{160}}{{40}}\]
Number of moles of NaOH = 4 moles
Hence, the number of moles of sodium hydroxide contained in 160 g of it is 4 moles.
Note: Mole concept simplifies the mass relation among reactants and products such that we can base our calculation on the coefficients (numbers of molecules involved in the reaction). At the same time, mass or the quantity of substance is on lab scale in grams.
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