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1 year ago

Classical bonding in NaH2PO4

Chemistry

1 answer:

lesantik [10]1 year ago

7 0

$NaH_2PO_4$ , a crystal structure with a short symmetrical hydrogen bond.

<h3>What is Classical bonding?</h3>

Classical models of the chemical bond. By classical, we mean models that do not take into account the quantum behaviour of small particles, notably the electron. These models generally assume that electrons and ions behave as point charges which attract and repel according to the laws of electrostatics.

Sodium dihydrogen phosphate is a derivative composed of glycerol derivatives formed by reacting mono and diglycerides that are derived from edible sources with phosphorus pentoxide followed by neutralization with sodium carbonate.

Bonding in $NaH_2PO_4$

$NaH_2PO_4$ , a crystal structure with a short symmetrical hydrogen bond. Sodium dihydrogen phosphate ( $NaH_2PO_4$ ) is monoclinic, space group P2,/c, with a= 6.808 (2), b= 13.491 (3), c=7.331 (2)/~, fl=92.88 (3) ; Z=8.

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Explanation:

We are given:

Equal masses of methanol $CH_4O$ and ethanol $C_2H_6O$ are mixed.

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Calculating the moles of methanol in the solution, by using the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}=\frac{xg}{32.04g/mol}$

Calculating the moles of ethanol in the solution, by using the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}=\frac{xg}{46.07g/mol}$

To calculate the mole fraction of methanol, we use the equation:

$\chi_{methanol}=\frac{n_{methanol}}{n_{methaol}+n_{ethanol}}$

$\chi_{methanol}=\frac{\frac{xg}{32.04g/mol}}{\frac{xg}{32.04g/mol}+\frac{xg}{46.07g/mol}}=0.590$

To calculate the mole fraction of ethanol, we use the equation:

$\chi_{ethanol}=\frac{n_{ethanol}}{n_{methaol}+n_{ethanol}}$

$\chi_{ethanol}=\frac{\frac{xg}{46.07g/mol}}{\frac{xg}{32.04g/mol}+\frac{xg}{46.07g/mol}}=0.410$

Thus mole fraction of methanol is 0.590 and mole fraction of ethanol 0.410 in three significant figures.

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