First you have to find the number of moles , then you have to apply stoichometry to find the number of moles of H2 gas , after that you can determine its mass.
Answer:
50 mg.
Explanation:
From the prescription, 240 ml amount is to be prepared containing 30 ml tussin syrup. As the frequency is 5 mL qid, it means 5mL of syrup is to be administered four times daily i.e. 20 mL daily. To calculate the amount of tussin taken daily,
<em> </em><em><u>30 mL tussin</u></em><em> = </em><em><u>X mL of tussin</u></em>
<em> 240 mL solution 20ml solution</em>
⇒ 2.5 mL of tussin
5 ml tussin contains 100mg guafenesin,
So,
⇒ 2.5 ml of tussin contains 100x2.5/5 = 50 mg guaifenesin
An oxide of nitrogen contains 30.45 mass % N, if the molar mass is 90± 5 g/mol the molecular formula is N₂O₄.
<h3>What is molar mass?</h3>
The molar mass of a chemical compound is determined by dividing its mass by the quantity of that compound, expressed as the number of moles in the sample, measured in moles. A substance's molar mass is one of its properties. The compound's molar mass is an average over numerous samples, which frequently have different masses because of isotopes.
<h3>How to find the molecular formula?</h3>
The whole-number multiple is defined as follows.
Whole-number multiple = 
The empirical formula mass is shown below.
Mw of empirical formula = Mw of N+ 2 x (Mw of O)
= 14.01 g/mol + 2 x (16.00 g/mol)
= 46.01 g/mol
With the given molar mass or the molecular formula mass, we can get the whole-number multiple for the compound.
Whole-number multiple = 
≈ 2
Multiplying the subscripts of NO2 by 2, the molecular formula is N(1x2)O(2x2)= N2O4.
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Well, clearly the calculated value for the number of hydrating water molecules would increase above its true level, because the total weight loss would be greater than expected. This is of course undesirable, but may usually be avoided by careful application of the experimental procedures. The signs to look for include
<span>(a) loss of water of hydration usually occurs at a considerably lower temperature than decomposition of the salt, because the water molecules are not strongly bonded in the hydrated complex. Dehydration typically occurs in a broad range of temperatures, typically from 50°C to around 200°C, whereas decomposition of the dehydrated salt generally takes place at temperatures over 200°C and in some case over 1000°C. So dehydration should be performed with care - avoid over-heating the sample in order to ensure that all the water has been driven off. </span>
<span>(b) dehydration often results in a change of appearance of the sample, particularly the colour and particle size of crystalline hydrates. However, decomposition may be accompanied by an additional change at higher temperatures, which gives a warning of its occurrence. </span>
<span>(c) if it is suspected that decomposition is occurring, or that dehydration is not complete, exploratory runs of varying duration at a given temperature may be carried out. There are two criteria to judge the effectiveness of the procedure </span>
<span>(i) the weight of the sample decreases to a constant stable value: this is a sign that dehydration is complete and decomposition - which is usually a much slower process - is not occurring. </span>
<span>(ii) the calculated number of molecules of water lost should take an integer value. If it differs by more than, say, 0.1 from an integer than it is probable that one of these two undesirable effects is present. Some hydrates lose water in steps through intermediate compounds with a lower level of hydration. These may provide plateaus where the weight loss is stable but dehydration is not complete. These will, in general, not provide an integer value for the number of water molecules present (because the calculation is based on the assumption that the residual sample is completely dehydrated salt).</span>
